Hi Wei Yang
>On Sun, Jun 09, 2019 at 05:10:28PM +0800, ChenGang wrote:
>>Usually the value of min_free_kbytes is multiply of 4, and in this case
>>,the right shift is ok.
>>But if it's not, the right-shifting operation will lose the low 2 bits,
>But PAGE_SHIFT is not always 12.
You are right, and this is not the key point, this is just an example.
>>and this cause kernel don't reserve enough memory.
>>So it's necessary to align the value of min_free_kbytes to multiply of 4.
>>For example, if min_free_kbytes is 64, then should keep 16 pages, but
>>if min_free_kbytes is 65 or 66, then should keep 17 pages.
>>
>>Signed-off-by: ChenGang <cg.chen@xxxxxxxxxx>
>>---
>> mm/page_alloc.c | 3 ++-
>> 1 file changed, 2 insertions(+), 1 deletion(-)
>>
>>diff --git a/mm/page_alloc.c b/mm/page_alloc.c index d66bc8a..1baeeba
>>100644
>>--- a/mm/page_alloc.c
>>+++ b/mm/page_alloc.c
>>@@ -7611,7 +7611,8 @@ static void setup_per_zone_lowmem_reserve(void)
>>
>> static void __setup_per_zone_wmarks(void) {
>>- unsigned long pages_min = min_free_kbytes >> (PAGE_SHIFT - 10);
>>+ unsigned long pages_min =
>>+ (PAGE_ALIGN(min_free_kbytes * 1024) / 1024) >> (PAGE_SHIFT - 10);
>In my mind, pages_min is an estimated value. Do we need to be so precise?
This is the key point, user can set this value through interface/proc/sys/vm/min_free_kbytes, so a bit more precise is better.
>> unsigned long lowmem_pages = 0;
>> struct zone *zone;
>> unsigned long flags;
>>--
>>1.8.5.6
>--
>Wei Yang
>Help you, Help me
